Question: You have found the following ages (in years) of all 4 gorillas at your local zoo: $ 20,\enspace 8,\enspace 22,\enspace 7$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{20 + 8 + 22 + 7}{{4}} = {14.3\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $20$ years $5.7$ years $32.49$ years $^2$ $8$ years $-6.3$ years $39.69$ years $^2$ $22$ years $7.7$ years $59.29$ years $^2$ $7$ years $-7.3$ years $53.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{32.49} + {39.69} + {59.29} + {53.29}} {{4}} $ $ {\sigma^2} = \dfrac{{184.76}}{{4}} = {46.19\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{46.19\text{ years}^2}} = {6.8\text{ years}} $ The average gorilla at the zoo is 14.3 years old. There is a standard deviation of 6.8 years.